Answer:
The answer is in 3 point
Choose which point you like
The electric output of the plant is 48.19 MW
First we need to calculate the water power, it is given by the formula
WP=ρQgh
Here, ρ=1000 kg/m3 is density of water,Q is the flow rate, g is the gravity, and h is the water head
Therefore, WP=1000*65*9.81*90=57388500 W=57.38 MW
Now the overall efficiency of the hydroelectric power plant is given as
η=
Plugging the values in the above equation
0.84=EP/57.38
EP=48.19 MW
Therefore, the electric output of the plant is 48.19 MW.
Answer:
What are we supposed to find, if it is kinetic energy then this is the solution.
K.E=1/2mv^2
K.E= kinetic energy
M=mass
V=velocity
K.E =0.5*55*0.6^2
K.E=9.9J
Explanation:
Answer:
v = -v₀ / 2
Explanation:
For this exercise let's use kinematics relations.
Let's use the initial conditions to find the acceleration of the electron
v² = v₀² - 2a y
when the initial velocity is vo it reaches just the negative plate so v = 0
a = v₀² / 2y
now they tell us that the initial velocity is half
v’² = v₀’² - 2 a y’
v₀ ’= v₀ / 2
at the point where turn v = 0
0 = v₀² /4 - 2 a y '
v₀² /4 = 2 (v₀² / 2y) y’
y = 4 y'
y ’= y / 4
We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.
v² = v₀² -2a y’
v² = 0 - 2 (v₀² / 2y) y / 4
v² = -v₀² / 4
v = -v₀ / 2
We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.
<u>Answer:</u>
Work input = Work output * Work against friction is your answer so C
<u>Explanation:</u>
I hope this helps you :)